Question

在查看Control.Monad文档后，我感到很困惑这段话：

上述法律意味着：

fmap f xs = xs >>= return . f

他们如何暗示？

Answer 1

Control.Applicative说

作为这些法律的结果，f的Functor实例将满足
fmap f x = pure f <*> x

Applicative和Monad之间的关系表示

pure = return

(<*>) = ap

ap说

return f `ap` x1 `ap` ... `ap` xn

相当于

liftMn f x1 x2 ... xn

因此

fmap f x = pure f <*> x
         = return f `ap` x
         = liftM f x
         = do { v <- x; return (f v) }
         = x >>= return . f

Answer 2

Functor instances are unique，从某种意义上说，如果F是Functor并且你有一个foobar :: (a -> b) -> F a -> F b函数，那么foobar id = id（也就是说，它跟随着foobar = fmap第一个仿函法）然后liftM :: Monad f => (a -> b) -> f a -> f b liftM f xs = xs >>= return . f。现在，考虑这个功能：

liftM id xs

那么liftM id xs xs >>= return . id -- id does nothing, so... xs >>= return -- By the second monad law... xs是什么？

liftM id xs = xs

liftM id = id;也就是liftM = fmap。因此，fmap f xs = xs >>= return . f;或者，换句话说......

Applicative

通过{{1}}法律路由的

epheriment's answer也是达成此结论的有效方式。

fmap和bind之间的关系

2 个答案: